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Chapter 3 - CHAPTER 3 Second Order Linear Dierential...

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CHAPTER 3 Second Order Linear Differential Equations 3.1 Introduction; Basic Terminology and Results Any second order differential equation can be written as F ( x,y,y,y ) = 0 This chapter is concerned with special yet very important second order equations, namely linear equations. Recall that a first order linear differential equation is an equation which can be written in the form y + p ( x ) y = q ( x ) where p and q are continuous functions on some interval I . A second order, linear differential equation has an analogous form. DEFINITION 1. A second order linear differential equation is an equation which can be written in the form y + p ( x ) y + q ( x ) y = f ( x ) (1) where p, q , and f are continuous functions on some interval I . The functions p and q are called the coefficients of the equation; the function f on the right-hand side is called the forcing function or the nonhomogeneous term . The term “forcing function” comes from applications of second-order linear equations; the description “nonhomogeneous” is given below. A second order equation which is not linear is said to be nonlinear . Examples (a) y - 5 y + 6 y = 3 cos 2 x . Here p ( x ) = - 5 , q ( x ) = 6 , f ( x ) = 3 cos 2 x are continuous functions on ( -∞ , ). (b) x 2 y - 2 xy + 2 y = 0. This equation is linear because it can be written in the form (1) as y - 2 x y + 2 x 2 y = 0 where p ( x ) = 2 /x, q ( x ) = 2 /x 2 , f ( x ) = 0 are continuous on any interval that does not contain x = 0. For example, we could take I = (0 , ). 63
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(c) y + xy 2 y - y 3 = e xy is a nonlinear equation; this equation cannot be written in the form (1). Remarks on “Linear.” Intuitively, a second order differential equation is linear if y appears in the equation with exponent 1 only, and if either or both of y and y appear in the equation, then they do so with exponent 1 only. Also, there are no so-called “cross- product” terms, yy,yy ,y y . In this sense, it is easy to see that the equations in (a) and (b) are linear, and the equation in (c) is nonlinear. Set L [ y ] = y + p ( x ) y + q ( x ) y . If we view L as an “operator” that transforms a twice differentiable function y = y ( x ) into the continuous function L [ y ( x )] = y ( x ) + p ( x ) y ( x ) + q ( x ) y ( x ) , then, for any two twice differentiable functions y 1 ( x ) and y 2 ( x ), L [ y 1 ( x ) + y 2 ( x )] = [ y 1 ( x ) + y 2 ( x )] + p ( x )[ y 1 ( x ) + y 2 ( x )] + q ( x )[ y 1 ( x ) + y 2 ( x )] = y 1 ( x ) + y 2 ( x ) + p ( x )[ y 1 ( x ) + y 2 ( x )] + q ( x )[ y 1 ( x ) + y 2 ( x )] = y 1 ( x ) + p ( x ) y 1 ( x ) + q ( x ) y 1 ( x ) + y 2 ( x ) + p ( x ) y 2 ( x ) + q ( x ) + y 2 ( x ) = L [ y 1 ( x )] + L [ y 2 ( x )] and, for any constant c , L [ cy ( x )] = [ cy ( x )] + p ( x )[ cy ( x )] + q ( x )[ cy ( x )] = cy ( x ) + p ( x )[ cy ( x )] + cq ( x ) y ( x ) = c [ y ( x ) + p ( x ) y ( x ) + q ( x ) y ( x )] = cL [ y ( x )] . Therefore, as introduced in Section 2.1, L is a linear differential operator. This is the real reason that equation (1) is said to be a linear differential equation. The first thing we need to know is that an initial-value problem has a solution, and that it is unique. THEOREM 1. (Existence and Uniqueness Theorem) Given the second order linear equation (1). Let a be any point on the interval I , and let α and β be any two real numbers. Then the initial-value problem y + p ( x ) y + q ( x ) y = f ( x ) , y ( a ) = α, y ( a ) = β has a unique solution.
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